Question

A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move into a region of higher potential or lower potential? higher potential lower potential Previous Answers Correct Part BPart complete What was the potential difference that stopped the proton? Express your answer with the appropriate units.

Answer 1

Answer:

(a) Region of higher potential

(b) ΔV=1.911V

Explanation:

For Part (a)

Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest

For Part (b)

Potential difference we can calculate bu definition of kinetic energy

ΔK=mv²/2= -qΔV

express in ΔV

ΔV= -mv²/2q

Substitute the given values

So

ΔV= -mv²/2q

$=-\frac{mv^2}{2e}\\ =\frac{9.11*10^{-31}*(8.20*10^{5}m/s)^2}{2*1.602*10^{-19}}\\ =1.911V$

ΔV=1.911V